2

以下情况：

我将一些值保存到我的应用程序设置（Properties.Settings）中。在此设置中，我还保存了一些对象，这些对象的属性存储在其中。

现在我想实现一个函数来将我的设置导出到一个 xml 文件中。

对于整数值和字符串，我是这样做的：

XmlWriter writer = XmlWriter.Create(path, settings);
writer.WriteStartDocument();
writer.WriteElementString("ServerIP", Settings.Default.ServerIP);
writer.WriteElementString("ServerPort", Settings.Default.ServerPort.ToString());
writer.WriteEndDocument();
writer.Flush();
writer.Close();

好的..我希望这是直到现在的正确方法。

现在我需要将我的对象存储到这个文件中。

我的想法是使用XmlSerializer类。但不幸的是，我完全不知道如何使用它来将对象与一个 XML 文件中的其他值结合起来

此外，这里是我想写入 XML 文件的类的代码：http: //pastebin.com/PmB4tM7b

4

1 回答 1

1

为了对您的对象使用 XML 序列化，您需要使用预定义的序列化属性来装饰包含您的类的类或数据结构：

// block of settings like 
// <Service>
//  <Name>Service1</Name>
//  <Description>Starts the service 1</Description>
// </Service>
public class SettingsService
{
    // will be a node in the XML file
    [XmlElement(ElementName="Name")]
    public string Name { get; set; }
    // will be a node too
    [XmlElement(ElementName="Description")]
    public string Description { get; set; }
}

// holds a list of services
// <Services>
//  <Service>...</Service>
//  <Service>...</Service>
// </Services>
public class ServicesSettings
{
    // list of services
    [XmlArray(ElementName="SettingsServices")]
    public List<SettingsService> Services { get; set; }
    // single value!
    [XmlElement(ElementName="SettingsPort")]
    public int PortNumber { get; set; }
}

// serializes the objects to XML file
public void SerializeModels(string filename, ServicesSettings settings)
{
    var xmls = new XmlSerializer(settings.GetType());
    var writer = new StreamWriter(filename);
    xmls.Serialize(writer, settings);
    writer.Close();
}

// retrieves the objects from an XML file
public ServicesSettings DeserializeModels(string filename)
{
    var fs = new FileStream(filename, FileMode.Open);
    var xmls = new XmlSerializer(typeof(WindowsServicesControllerSettings));
    return (WindowsServicesControllerSettings) xmls.Deserialize(fs);
}

可以这样使用：

var service1 = new SettingsService { Name = "Service1", Description = "Blah blah" };
var service2 = new SettingsService { Name = "Service2", Description = "Blah blah blup" };
var services = new List<SettingsService> { service1, service2 };
var settings = new ServicesSettings
{ 
    Services = services,
    PortNumber = 1234
};

this.SerializeModels(@"d:\temp\settings.xml", settings);

此行创建以下 XML 文件：

<?xml version="1.0" encoding="utf-8"?>
<ServicesSettings xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
  <SettingsServices>
    <SettingsService>
      <Name>Service1</Name>
      <Description>Blah blah</Description>
    </SettingsService>
    <SettingsService>
      <Name>Service2</Name>
      <Description>Blah blah blup</Description>
    </SettingsService>
  </SettingsServices>
  <SettingsPort>1234</SettingsPort>
</ServicesSettings>

从 XML 文件中检索对象：

var settings = this.DeserializeModels(@"d:\temp\settings.xml");
于 2013-09-08T19:30:10.003 回答